A) \[{{\sec }^{2}}x\]
B) \[-\text{cose}{{\text{c}}^{2}}x\]
C) \[2\,{{\sec }^{2}}\frac{x}{2}\]
D) \[-2\text{cose}{{\text{c}}^{2}}\frac{x}{2}\]
Correct Answer: B
Solution :
\[\frac{d}{dx}\sqrt{\frac{1+\cos 2x}{1-\cos 2x}}=\frac{d}{dx}\cot x=-\text{cose}{{\text{c}}^{2}}x\].
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